The domain of the function `f(x)=(1)/(log_(10)(1-x))+sqrt(x+2)` is equal to

Options

A. `[-3,-2.5)cup(-2.5,2]`
B. `[-2,0)cup(0,1]`
C. `[0,1]`
D. None of these

Answer

To find the domain of the function \( f(x) = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2} \), we need to ensure that both components of the function are defined.

### Step 1: Analyze the logarithmic part
The logarithmic function \( \log_{10}(1-x) \) is defined when its argument is positive and not equal to zero. Thus, we have two conditions to consider:

1. **Condition for the logarithm to be defined**:
\[ 1 - x > 0 \implies x < 1 \]

2. **Condition for the logarithm to not be zero**:
\[ \log_{10}(1-x) \neq 0 \implies 1 - x \neq 10^0 \implies 1 - x \neq 1 \implies x \neq 0 \]

### Step 2: Analyze the square root part
The square root function \( \sqrt{x + 2} \) is defined when its argument is non-negative:
\[ x + 2 \geq 0 \implies x \geq -2 \]

### Step 3: Combine the conditions
Now we need to combine all the conditions derived:

1. From the logarithm, we have:
- \( x < 1 \)
- \( x \neq 0 \)

2. From the square root, we have:
- \( x \geq -2 \)

### Step 4: Determine the intervals
Now, we can summarize the conditions:
- The values of \( x \) must be greater than or equal to \(-2\) and less than \(1\), excluding \(0\).

Thus, the valid intervals are:
- From \(-2\) to \(0\) (inclusive of \(-2\) but exclusive of \(0\)): \([-2, 0)\)
- From \(0\) to \(1\) (exclusive of \(0\) and exclusive of \(1\)): \((0, 1)\)

### Final Domain
Combining these intervals, the domain of the function \( f(x) \) is:
\[ [-2, 0) \cup (0, 1) \]